1+1<or>or=2 proof that it>or<2?

Answer 1

proof :

1 -1 = 1 - 1
lhs: 1 - 1
rhs: 1 -1

thaking 1 common from lhs v hav
1 (1 - 1 )

expanding rhs by the formula a^2 - b^2 = (a+b)(a-b)

(1-1)(1+1)

thus we hav

1 (1 - 1 ) = ( 1 - 1 ) (1 + 1 )
dividing througout by ( 1 - 1 )

we get 1 = 1 + 1



Does this help u??

Answer 2

You know what are significant numbers.I will tell you.If there is any number say 1.66,while considering it in significant number we will take it as 2.Actually anything > .5 is taken as the next number(as above 1.66 is taken as 2).On the contrary anything < .5 is taken as the previous number.Understand this and you will get your answer.Here,I would like to tell you that your problem is that of Inequality and the procedure of its solution may not satisify you completely.

Answer 3

given a and b are both real numbers, only one of the following statements is true: ab.
I'm going with 1+ 1 = 10. in binary.

Answer 4

You can prove (wrongly) that 1=.99999...
and thus 1+1=.99999.... + .9999999... = 1.9999999....<2

Answer 5

when count wrongly it is proved u get > or < 2

Answer 6

the previous poster is wrong: 0.9999... does = 1.000... and in fact 1.9999... does = 2.